Let \(f(z)\) be an analytic function on \(|z|<1\). Prove that \(f(z)\) is necessarily a constant if \(f(\bar{z})\) is also analytic.
Solution
Let \(\tilde f(z) \da f(\bar z)\). Using that \(f\) is analytic iff its components solve Cauchy-Riemann, using that \(f, \tilde f\) are analytic, \[ u_x = v_y && u_y = -v_x \\ u_x = -v_y && u_y = v_x \\ \\ \implies 2u_x = v_y - v_y = 0 \implies u_x = 0 \\ \implies 2u_y = v_x - v_x = 0 \implies u_y = 0 \\ \implies 0 = u_y - u_y = v_x - (-v_x) = 2v_x \implies v_x = 0 \\ \implies 0 = u_x - u_x = v_y - (-v_y) = 2v_y \implies v_y = 0 ,\] so \(\grad u = [u_x, u_y] \equiv \vector 0\) making \(u\) constant. Similarly \(\grad v = [v_x, v_y] = \vector 0\), so \(f: \RR^2\to \RR\) is constant.